Let's Play Chess on Other Planet

Consider a n × n n\times n chess board. Let the total number of possible rectangles and squares be R n { R }_{ n } and S n { S }_{ n } respectively.

lim n R n n S n \displaystyle{\lim _{ n\rightarrow \infty }{ \cfrac { { R }_{ n } }{ { n \ S }_{ n } } }}

If the limit above is in the form of a b \frac a b for coprime positive integers a , b a,b , find a + b a+b .

Details and Assumptions

  • All squares are rectangles, but not all rectangles are squares.
Original


The answer is 7.

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2 solutions

Discussions for this problem are now closed

Deepanshu Gupta
Jan 17, 2015

R n = ( n + 1 2 ) ( n + 1 2 ) = n 2 ( n + 1 ) 2 4 S n = 1 n r 2 = n ( n + 1 ) ( 2 n + 1 ) 6 L = lim n 3 ( n + 1 ) 2 ( 2 n + 1 ) = lim n 3 ( 1 + 1 n ) 2 ( 2 + 1 n ) = 3 4 \displaystyle{\\ { R }_{ n }=\left( \begin{matrix} n+1 \\ 2 \end{matrix} \right) \left( \begin{matrix} n+1 \\ 2 \end{matrix} \right) =\cfrac { { n }^{ 2 }{ (n+1) }^{ 2 } }{ 4 } \\ { S }_{ n }=\sum _{ 1 }^{ n }{ { r }^{ 2 } } =\cfrac { n(n+1)(2n+1) }{ 6 } \\ L=\lim _{ n\rightarrow \infty }{ \cfrac { 3(n+1) }{ 2(2n+1) } } =\lim _{ n\rightarrow \infty }{ \cfrac { 3(1+\cfrac { 1 }{ n } ) }{ 2(2+\cfrac { 1 }{ n } ) } } =\cfrac { 3 }{ 4 } }

Here For calculating Rectangles There are n+1 Horizontal lines and n+1 vertical Lines so we select any two Vertical and any two horizontal Lines So that Rectangle is formed (and obviously squares are also rectangles)

You may Take Eg as Our Planet Chess Board ,(8*8) in This 9 are vertical and 9 horizontal lines are present.

Karan Keep it up bro ! Keep Posting such more :)

Thanks buddy !

Karan Shekhawat - 6 years, 4 months ago

How do you get Rn? Can you explain it?

Bhaskar Sukulbrahman - 6 years, 4 months ago

I add Few more Lines To solution , Hope U will get it .

Deepanshu Gupta - 6 years, 4 months ago
Bhargav Upadhyay
Feb 4, 2015

Solution is here..

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